We’re ready to start proving the Harris-Kesten theorem, so from now on, everything we say is about bond percolation in . The global structure of the argument is in three steps:
- Show that (this implies )
- Show that for , there is an infinite open cluster with probability 1 (this implies )
- Show that for (this implies that )
- Conclude that
This post will accomplish the first step.
1. Crossing Probabilities
We start with some more definitions: a rectangle in is a set of sites of the form . We say this rectangle has size , where . It contains sites and bonds.
This rectangle has two duals: a horizontal dual
and a vertical dual
We have that is a rectangle and is a rectangle. Additionally,
Finally, we define an open horizontal crossing of a rectangle to be an open path in connecting a site to a site , and similarly for an open vertical crossing. We write and for the events that such crossings exist.
The following lemma is, according to Bollobas, the “reason” why . It is intuitively clear, but working through all the details is a bit of slog, so let’s just take it at face value.
Lemma 1. Let be a rectangle in or its dual. Then exactly one of and holds.
Why exactly is this lemma the “reason” cited above? The following corollary sheds some light on that.
Corollary 2. If and are and rectangles in , then
Proof: The bonds of the dual are open when the bonds of the original graph are closed and vice versa. So by Lemma 1, each configuration of bonds will satisfy exactly one of and , so
But is indistinguishable from , so the result follows.
Now if we take in this corollary, we get that for an rectangle ,
But because the rectangle has dimension , the event has the same probability as , so they are both equal to .
So the intuition is somehow that when , it’s more likely you have a horizontal crossing than not, so you get an infinite open cluster. When , it’s less likely than not, so you don’t. The critical behavior is at exactly .
More importantly, we get that if is an square,
where we have the inequality because this event must be more likely than the crossing for an rectangle. This is a bound on the crossing probabilities of a square. We now have to extend this result to rectangles.
2. Squares to Rectangles
This next lemma relates crossings of squares to crossings of rectangles. We use the notation .
Lemma 3. Consider , an rectangle with . Let be the event that there are open paths such that crosses the square from top to bottom and lies within and joins some site on to the right hand side of . Then
Proof: Suppose occurs, so an open path crosses from top to bottom. Such a path cuts into two pieces, one on the left and on the right. Let be the left-most open vertical crossing (when one exists). For any , the event does not depend on bonds to the right of .
The key claim is that
Since is a disjoint union of events of the form , this implies that
which rearranges to
But occurring means that does as well, so , and the lemma is proven.
For the claim, let be the path (that may or may not be open) formed by the union of and it’s reflection in the horizontal axis of , with a bond joining to . This path is a vertical crossing of the rectangle .
Now, there is an open path crossing horizontally with probability . This path must meet , and the probability it does so at a site of is at least . Thus the event that there is an open path in to the right of joining a site of to the right end of has probability at least . This event depends only bonds to the right of , and all such bonds in are to the right of . These bonds are independent of , so
If and both occur, then does as well. So
proving the claim and thus the lemma.
Remark 1. If we instead take an rectangle and an rectangle , this proof will show that
and we will use this formulation later on.
(This argument is much simpler than it seems – drawing the picture goes a long way.)
3. Thinning the Rectangles Out
With this lemma, we can now bound the crossing probabilities for non-square rectangles. In particular, we look at a rectangle. For convenience, let , and since it’s the case we really care about, .
Corollary 4. We have .
Proof: Take two square rectangles that overlap in an strip. This strip contains two squares; call the smaller of these. (Again, draw the picture!) Let be the event described above, but reflected horizontally. Then we apply Lemma 4 to (which is coincidentally a square) to get
Now, we call back to Harris’s lemma from the previous post: the events are all increasingly and positive correlated. If they all hold, so does , and so we have
This is then
But by our observation at the end of Section 1, this is at least
A rectangle isn’t quite thing enough for our purposes, but the rest of the thinning can be done using the remark after Lemma 3. We have
and this is the bound that our proof of will hinge on.
4. Harris’s Theorem
Theorem 5. For bond percolation on , .
The actual result we’ll show is that
for a constant , and the stated result follows from this.
When , bonds in the dual lattice are open with probability , so by the computation above, the probability a rectangle has a crossing the long way is at least . Consider 4 such rectangles oriented to form an annulus around the origin, with the “hole” in the annulus being a square centerd at the origin.
By Harris’s lemma, the probability each of these rectangles is crossed the long way by an open path in the dual is at least . The union of these paths is an open dual cycle around the annulus.
Let be the square annulus centered on with inner radius and outer radius . Let be the event that contains an open dual cycle surrounding the inside of , and thus the origin.
By our argument above, for each . All of the are independent, and if holds, then there is no path from inside to outside , so
which is the desired bound.